Question

I sent values through file_get_contents.
My question is i am unable receive(print) GET method values in work.php.
I am using stream_context_create() this will create a resource id.

page name sendvalues.php

// Create a stream
$opts = array(
   http =>array(
     method =>"GET",
     phone  => "9848509317",
     msg    => "hi naveen"
  )
);

echo $context = stream_context_create($opts);

$file = file_get_contents( http://www.aakrutisolutions.com/projects/testingsite/smstest/sms_http_curl/work.php , false, $context);
echo $file;

page name work.php
echo "";
print_r($_GET);  /// i am unable to get my query string values
echo "";

Answer 1

仅附上你在座标上的ur:

$file = file_get_contents( http://www.aakrutisolutions.com/projects/testingsite/smstest/sms_http_curl/work.php?phone=123&msg=hi%20naveen );

The context options you used... are context options. Here is specified which you can user for http: http://www.php.net/manual/de/context.http.php It s not to be transmitted if you put random stuff in there.

Answer 2

Your array is incorrect. Look at the examples at http://php.net/manual/en/function.file-get-contents.php. You cannot simply add POST arguments to the context array.

提出《长期合作行动倡议》要求的正确背景阵列就是这样:

$opts = array(
   http  => array(
     method  =>  POST ,
     content  => http_build_query(array(
       phone  => 9848509317,
       msg    =>  hi naveen 
    ))
  )
);

www.un.org/Depts/DGACM/index_spanish.htm 或者简单地使用GET(如你在其他文字中期望的那样),这样就可以在URL中提出论点(与)。

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